Peter wrote: An integer-sided triangle has angles $ p\theta$ and $ q\theta$, where $ p$ and $ q$ are relatively prime integers. Prove that $ \cos\theta$ is irrational. The reference for this problems is "CRUX(No. 1. Vol. 24, 1998), Problem 2305, Richard I. Hess".

Added. Now all we need is a solution.

What about equilateral triangle? It can be $p=q=1, \theta = \frac{\pi}{3}$ It is integer-sided, but $\cos \frac{\pi}{3} = \frac{1}{2} \in \mathbb{Q}$, Then this can disprove the proposition.

maybe is: "p and q are distinct primes" instead of: "where p and q are relatively prime integers"

Unless I'm mistaken, I have a proof that $\cos \theta$ is rational. Let $S = \{n \in \mathbb{N} \ | \ \cos n\theta \in \mathbb{Q} \}$. Using that the angles of the triangle are $p\theta, \ q\theta, \ \pi - (p+q) \theta$, and the cosine rule, we have that $p,q,p+q \in S$. Now, $\cos |(x-y)| \theta = 2 \cos x\theta \cdot \cos y\theta - \cos (x+y)\theta$, means that $x,y,x+y \in S \implies |x-y| \in S$. But this means that if $x,y,x+y \in S$, then gcd$(x,y) \in S$. But, then since $p,q$ are coprime, so $1 \in S$.